lebesgue integral - Exercise in finite measure space.

Let $(X,\mu)$ be a finite measure space and $f:X\to[0,\infty]$ be a measurable function. Define a function $\varphi:\mathbb{R}^+\to\mathbb{R}^+$ by $\varphi(r)=\left(\frac{1}{\mu(X)}\int_Xf^rd\mu\right)^\frac1r$. Prove that $\varphi(r)\leq\varphi(s)$ if $r

Answer

We have
$$\varphi(r) = \left(\frac1{\mu(X)}\int_X f^r\ \mathsf d\mu\right)^{\frac1r} = \left(\int_X f^r \ \frac1{\mu(X)}\ \mathsf d\mu \right)^{\frac1r}.$$
Now, by Hölder's inequality, in general, for measurable functions $F, G$ and conjugate exponents $p,q$, that is, $\frac1p + \frac1q = 1$ where $1 $$\begin{align} \phi(r) &= \left(\int_X FG\ \mathsf d\mu\right)^{\frac1r}\\ &\leqslant \left(\int_X F^{\frac sr}\ \mathsf d\mu\right)^{\frac r{rs}}\left(\int_X G^{\frac s{s-r}} \right)^{\frac{s-r}{rs}}\\ &=\left(\int_X f^s\ \mathsf d\mu \right)^{\frac 1s}\left(\int_X \left(\frac1{\mu(X)}\right)^{\frac s{s-r}} \ \mathsf dx\right)^{\frac{s-r}{rs}}\\ &=\left(\int_X f^s\ \mathsf d\mu \right)^{\frac 1s}\left(\mu(X)^{\frac s{r-s}}\mu(X) \right)^{\frac{s-r}{rs}}\\ &=\left(\frac1{\mu(X)}\int_X f^s\ \mathsf d\mu \right)^{\frac1s}\\ &=\varphi(s), \end{align}$$
as desired.